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\begin{center}
\vskip 1cm{\LARGE\bf A Note on Sumsets and Restricted Sumsets
\vskip 1cm}
\large
Jagannath Bhanja\\
Harish-Chandra Research Institute\\ 
Chhatnag Road, Jhunsi\\ 
Prayagraj 211019\\ 
India \\
\href{mailto:jagannathbhanja@hri.res.in}{\tt jagannathbhanja@hri.res.in} \\
\end{center}
\vskip .2 in


\begin{abstract} 
In this note we find the optimal lower bound for the size of the sumsets $HA$ and $H\,\hat{}A$ over finite sets $H, A$ of nonnegative integers, where $HA = \bigcup_{h\in H} hA$ and $H\,\hat{}A = \bigcup_{h\in H} h\,\hat{}A$. We also find the underlying algebraic structure of the sets $A$ and $H$ for which the size of the sumsets $HA$ and $H\,\hat{}A$ is minimum.  
\end{abstract} 


\section{Introduction}\label{sec1}

For a given finite set $A$ of integers and for a positive integer $h$, the {\em sumset} $hA$ and the {\em restricted sumset} $h\,\hat{}A$ are fundamental objects in the field of \emph{additive number theory}. The sumset $hA$ is the set of integers that can be written as the sum of $h$ elements of $A$, whereas the sumset $h\,\hat{}A$ is the set of integers that can be written as the sum of $h$ 
{\it pairwise distinct\/} elements of $A$. In this regard, two of the important problems in additive number theory are to find the best possible lower bounds for the size of the sumsets $hA$ and $h\,\hat{}A$, and to find the structure of the finite set $A$ for which the sumsets $hA$ and $h\,\hat{}A$ contain the
minimum number of elements. These two problems have been well established in the group of integers \cite{nathu, nathanson}.

\begin{theorem} {\cite[Theorem 1.4, Theorem 1.6]{nathanson}} \label{theoremI}
	Let $A$ be a finite set of $k$ integers. Let $h$ be a positive integer. Then
	\begin{equation*}
	|hA|\geq h(k-1)+1.
	\end{equation*}
	Moreover, if this lower bound is exact with $h\geq 2$, then $A$ is an arithmetic progression.
\end{theorem}

\begin{theorem} {\cite[Theorem 1, Theorem 2]{nathu}} \label{theoremII}
	Let $A$ be a finite set of $k$ integers. Let $h\leq k$ be a positive integer. Then
	\begin{equation*}
	|h\,\hat{}A|\geq h(k-h)+1.
	\end{equation*}
	Moreover, if this lower bound is exact with $k\geq 5$ and $2\leq h\leq k-2$, then $A$ is an arithmetic progression.
\end{theorem}


Now let $H$ be a given finite set of nonnegative integers. Define the sumset \cite[p.\ 175]{bajnok}
\[HA := \bigcup_{h\in H} hA,\]
and the restricted sumset
\[H\,\hat{}A := \bigcup_{h\in H} h\,\hat{}A.\]
Here we are assuming that $0A=0\,\hat{}A=\{0\}$. 


For a set $A$ and for an integer $c$, we let $c \cdot A = \{ca: a \in A\}$. For integers $a, b$ with $a \leq b$, we also let $[a, b]=\{a,a+1,\ldots,b\}$.


The case $H=[0, r]$ is more interesting and has been studied before (for recent papers, see \cite{bajnok, balandraud, bhanja}). Bajnok \cite{bajnok} defined the sumsets for an arbitrary finite set $H$ of nonnegative integers, and asked to 
study similar problems like the sumsets $hA$ and $h\,\hat{}A$ over finite abelian groups. That is, find the optimal lower bound for the size of the sumsets $HA$ and $H\,\hat{}A$, and the structure of the sets $H$, $A$ for which the sumsets $HA$, $H\,\hat{}A$ contain the minimum number of elements.


In this note, we study these two problems for the sumset $HA$ in Section \ref{sec2}, and the sumset $H\,\hat{}A$ in Section \ref{sec3}, for finite sets $A$ of nonnegative integers (or nonpositive integers) and $H$ of nonnegative integers. We consider two separate cases, namely
\begin{itemize}
\item[(i)] the set $A$ consists of positive integers and
\item[(ii)] the set $A$ consists of nonnegative integers.
\end{itemize}
The cases
\begin{itemize}
\item[(iii)] the set $A$ consists of negative integers and
\item[(iv)] the set $A$ consists of nonpositive integers,
\end{itemize}
follow from the cases (i) and (ii), respectively, as $H(c \cdot A) = c \cdot HA$ and $H\,\hat{} (c \cdot A) = c \cdot H\,\hat{}A$ for arbitrary integers $c$. As consequences of our results we obtain some recent results in this direction.


In Section \ref{sec2} and Section \ref{sec3}, we use the following notation:
for a set $S=\{s_{1}, s_{2}, \ldots, s_{k-1}, s_{k}\}$ with $k \geq 2$ and $s_{1}&lt; s_{2}&lt; \cdots&lt; s_{k-1}&lt; s_{k}$, we write $\min(S)=s_{1}$, $\min\nolimits_{+}(S)=s_{2}$, $\max(S)=s_{k}$, and $\max\nolimits_{-}(S)=s_{k-1}$.  


\section{Regular sumset}\label{sec2}

\begin{theorem}\label{sumset-thm}
	Let $A$ be a set of $k$ positive integers. Let $H$ be a set of $r$ positive integers with $\max(H)=h_{r}$. Then
	\begin{equation}\label{sumset-eqn1}
	|HA| \geq h_{r}(k-1)+r.
	\end{equation}
	This lower bound is optimal.
\end{theorem}

\begin{proof}
	Let $A=\{a_{1},a_{2},\ldots,a_{k}\}$ and $H=\{h_{1},h_{2},\ldots,h_{r}\}$, where $0&lt;a_{1}&lt;a_{2}&lt;\cdots&lt;a_{k}$ and $0&lt;h_{1}&lt;h_{2}&lt;\cdots&lt;h_{r}$. Set
	\begin{equation}\label{set1}
	S_{1} := h_{1}A
	\end{equation}
	and
	\begin{equation}\label{seti}
	S_{i} := (h_{i}-h_{i-1})A+h_{i-1}a_{k},
	\end{equation}
	for $i=2,3,\ldots,r$. Clearly $S_{i} \subseteq h_{i}A$ for $i=1,2,\ldots,r$, and $\max(S_{i}) &lt; \min(S_{i+1})$ for $i=1,2,\ldots,r-1$. Therefore $S_{1}, S_{2}, \ldots, S_{r}$ are pairwise disjoint subsets of $HA$. Hence, by Theorem \ref{theoremI}, we have 
	\begin{align}\label{sumset-eqn2}
	|HA| &amp;\geq \sum_{i=1}^{r} |S_{i}|\nonumber\\
	&amp;= |h_{1}A|+\sum_{i=2}^{r} |S_{i}|\nonumber\\
	&amp;\geq h_{1}(k-1)+1+\sum_{i=2}^{r} [(h_{i}-h_{i-1})(k-1)+1]\nonumber\\
	&amp;= h_{r}(k-1)+r.
	\end{align}
	This proves (\ref{sumset-eqn1}).
	
	
	To see that the lower bound in (\ref{sumset-eqn1}) is optimal, let $A=[1, k]$ and $H=[1, r]$. Then $HA=[1, rk]$, and  hence $|HA|=rk$. This completes the proof of the theorem.
\end{proof}

\begin{remark}
	If $A$ contains nonnegative integers with $0 \in A$, then $HA=h_{r}A$, as $h_{i}A \subseteq h_{r}A$ for $i=1,2,\ldots,r-1$. Therefore, by Theorem \ref{theoremI}, we have $|HA| \geq h_{r}(k-1)+1$. Furthermore, this bound is optimal, and it can be seen by taking $A=[0, k-1]$ and $H=[1, r]$.
\end{remark} 

Now we prove the inverse result of Theorem \ref{sumset-thm}.

\begin{theorem}\label{sumset-inverse-thm}
	Let $A$ be a set of $k\geq 2$ positive integers and $H$ be a set of $r \geq 2$ positive integers with $\max(H)=h_{r}$. If $|HA|=h_{r}(k-1)+r$, then $H$ is an arithmetic progression of difference $d$ and $A$ is an arithmetic progression of difference $d \cdot \min(A)$.
\end{theorem}

\begin{proof}
	Let $A=\{a_{1},a_{2},\ldots,a_{k}\}$ and $H=\{h_{1},h_{2},\ldots,h_{r}\}$, where $0&lt;a_{1}&lt;a_{2}&lt;\cdots&lt;a_{k}$ and $0&lt;h_{1}&lt;h_{2}&lt;\cdots&lt;h_{r}$. Let $|HA|=h_{r}(k-1)+r$. Then the sumset $HA$ contains precisely the elements of the sets $S_{i}$ for $i=1, \ldots, r$, which are defined in (\ref{set1}), (\ref{seti}). 
	
	
	First, we show that $A$ is an arithmetic progression. Observe that the assumption $|HA|=h_{r}(k-1)+r$ together with (\ref{sumset-eqn2}) implies $|h_{1}A|=h_{1}(k-1)+1$. If $h_{1}&gt;1$, then from Theorem \ref{theoremI}, it follows that the set $A$ is an arithmetic progression. So, let $h_{1}=1$. Then
	\[S_{1}=h_{1}A=A=\{a_{1},a_{2},\ldots,a_{k}\}.\]
	Set \[S:=\{a_{1}, h_{2}a_{1}, (h_{2}-1)a_{1}+a_{2}, \ldots, (h_{2}-1)a_{1}+a_{k-1}\}.\]
	Clearly $S \subseteq HA$ and $\max(S)=(h_{2}-1)a_{1}+a_{k-1}&lt;(h_{2}-1)a_{1}+a_{k}=\min(S_{2})$. Thus $S=S_{1}$. In other words, $(h_{2}-1)a_{1}+a_{i-1}=a_{i}$ for $i=2,3,\ldots,k$. Equivalently, $a_{i}-a_{i-1}=(h_{2}-1)a_{1}$ for $i=2,3,\ldots,k$. Hence, $A$ is an arithmetic progression.
	
	
	Next we show that $H$ is an arithmetic progression. For $i=1,2, \ldots,r-1$, consider the integers $(h_{i+1}-h_{i})a_{1}+a_{k-1}+(h_{i}-1)a_{k}$. Clearly
	\begin{multline*}
	\max\nolimits_{-}(S_{i})=a_{k-1}+(h_{i}-1)a_{k}&lt;(h_{i+1}-h_{i})a_{1}+a_{k-1}+(h_{i}-1)a_{k}\\
	&lt;(h_{i+1}-h_{i})a_{1}+h_{i}a_{k}=\min(S_{i+1}).
	\end{multline*}
	But we already have
	\begin{align*}
	\max\nolimits_{-}(S_{i})=a_{k-1}+(h_{i}-1)a_{k}&lt;h_{i}a_{k}=\max(S_{i})&lt;(h_{i+1}-h_{i})a_{1}+h_{i}a_{k}=\min(S_{i+1}).
	\end{align*}		
	Thus
	\[(h_{i+1}-h_{i})a_{1}+a_{k-1}+(h_{i}-1)a_{k}=h_{i}a_{k} \ \text{for} \ i=1,2, \ldots,r-1.\]
	This implies
	\begin{equation}\label{sumset-eqn3}
	a_{k}-a_{k-1}=(h_{i+1}-h_{i})a_{1} \ \text{for} \ i=1,2, \ldots,r-1.
	\end{equation}
	Therefore
	\[h_{2}-h_{1} = h_{3}-h_{2} = \cdots = h_{r}-h_{r-1},\]
	and hence the set $H$ is an arithmetic progression. Furthermore, by (\ref{sumset-eqn3}), the set $A$ is an arithmetic progression of difference $(h_{i+1}-h_{i})a_{1}$. This completes the proof of the theorem.
\end{proof}


\section{Restricted sumset}\label{sec3}

\begin{theorem}\label{restricted-sumset-thm}
	Let $A$ be a set of $k$ positive integers and $H=\{h_{1}, h_{2}, \ldots, h_{r}\}$ be a set of positive integers with $h_{1}&lt;h_{2}&lt;\cdots&lt;h_{r} \leq k$. Set $h_{0}=0$. Then
	\begin{equation}\label{restricted-eqn1}
	|H\,\hat{}A| \geq \sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-h_{i})+r.
	\end{equation}
	This lower bound is optimal.
\end{theorem}

\begin{proof}
	Let $A=\{a_{1},a_{2},\ldots,a_{k}\}$, where $a_{1}&lt;a_{2}&lt;\cdots&lt;a_{k}$. Set
	\begin{equation}\label{reset1}
	S_{1} := h_{1}\,\hat{}A
	\end{equation}
	and
	\begin{equation}\label{reseti}
	S_{i} := (h_{i}-h_{i-1})\,\hat{}A_{i}+\max(h_{i-1}\,\hat{}A),
	\end{equation}
	for $i=2,3,\ldots,r$, where $A_{i}=\{a_{1},a_{2},\ldots,a_{k-h_{i-1}}\}$. Clearly $S_{i} \subseteq h_{i}\,\hat{}A$ for $i=1,2,\ldots,r$, and $\max(S_{i}) &lt; \min(S_{i+1})$ for $i=1,2,\ldots,r-1$. Therefore $S_{1}, S_{2}, \ldots, S_{r}$ are pairwise disjoint subsets of $H\,\hat{}A$. Hence, by Theorem \ref{theoremII}, we have
	\begin{align}\label{restricted-eqn2}
	|H\,\hat{}A| &amp;\geq \sum_{i=1}^{r} |S_{i}|\nonumber\\
	&amp;= |h_{1}\,\hat{}A|+\sum_{i=2}^{r} |S_{i}|\nonumber\\
	&amp;\geq h_{1}(k-h_{1})+1+\sum_{i=2}^{r} [(h_{i}-h_{i-1})(k-h_{i})+1]\nonumber\\
	&amp;= \sum_{i=1}^{r} (h_{i}-h_{i-1})(k-h_{i})+r.
	\end{align}
	This proves (\ref{restricted-eqn1}).
	
	
	Next to see that the lower bound in (\ref{restricted-eqn1}) is optimal, let $A=[1, k]$ and $H=[1, r]$ with $r \leq k$. Then $H\,\hat{}A \subseteq [1, k+(k-1)+\cdots+(k-r+1)]$. Therefore  $|H\,\hat{}A| \leq rk-\frac{r(r-1)}{2}$. This together with (\ref{restricted-eqn1}) implies $|H\,\hat{}A|=rk-\frac{r(r-1)}{2}$, and hence completes the proof of the theorem.
\end{proof}

As a consequence of Theorem \ref{restricted-sumset-thm}, we obtain the following corollary.

\begin{corollary}\label{restricted-sumset-nonve-cor}
	Let $A$ be a set of $k$ nonnegative integers with $0 \in A$. Let $H=\{h_{1}, h_{2}, \ldots, h_{r}\}$ be a set of positive integers with $h_{1}&lt;h_{2}&lt;\cdots&lt;h_{r} \leq k-1$. Set $h_{0}=0$. Then
	\begin{equation}\label{restricted-eqn3}
	|H\,\hat{}A| \geq \sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-h_{i}-1)+r+1.
	\end{equation}
	This lower bound is optimal.
\end{corollary}

\begin{proof}
	Let $A^{\prime}=A\setminus \{0\}$. Then $H\,\hat{}A^{\prime} \cup \{0\} \subseteq H\,\hat{}A$. This implies
	\[|H\,\hat{}A| \geq |H\,\hat{}A^{\prime}|+1 \geq \sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-h_{i}-1)+r+1.\]
	
	Furthermore, the optimality of the lower bound in (\ref{restricted-eqn3}) can be checked by taking $A=[0, k-1]$ and $H=[1, r]$, where $k, r$ are positive integers with $r \leq k-1$.
\end{proof}


The following result (which has recently been proved) is a particular case of Theorem \ref{restricted-sumset-thm} and Corollary \ref{restricted-sumset-nonve-cor}.

\begin{corollary} {\cite[Theorem 2.1, Corollary 2.1]{bhanja}}
	Let $A$ be a set of $k$ nonnegative integers and $H=[0, r]$ with $r \leq k$. If $0 \notin A$, then
	\begin{equation*}
	|H\,\hat{}A| \geq rk-\frac{r(r-1)}{2}+1.
	\end{equation*}
	If $0 \in A$ and $r \leq k-1$, then
	\begin{equation*}
	|H\,\hat{}A| \geq rk-\frac{r(r+1)}{2}+1.
	\end{equation*}
	These lower bounds are optimal. 	
\end{corollary}


Now we prove the inverse theorem of Theorem \ref{restricted-sumset-thm}.

\begin{theorem}\label{restricted-sumset-inverse-thm}
	Let $A$ be a set of $k\geq 6$ positive integers. Let $H=\{h_{1}, h_{2}, \ldots, h_{r}\}$ be a set of $r\geq 2$ positive integers with $h_{1}&lt;h_{2}&lt;\cdots&lt;h_{r}\leq k-1$. Set $h_{0}=0$. If
	\[|H\,\hat{}A|=\sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-h_{i})+r,\]
	then $H=h_{1}+[0, r-1]$ and $A = \min(A) \cdot [1, k]$.
\end{theorem}

\begin{proof}
	Let $A=\{a_{1},a_{2},\ldots,a_{k}\}$, where $0&lt;a_{1}&lt;a_{2}&lt;\cdots&lt;a_{k}$. Let $|H\,\hat{}A|=\sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-h_{i})+r$. Then the sumset $H\,\hat{}A$ contains precisely the elements of the sets $S_{i}$ for $i=1, \ldots, r$, which are defined in (\ref{reset1}), (\ref{reseti}).
	
	
	First, we show that $A$ is an arithmetic progression. Since $|H\,\hat{}A| = \sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-h_{i})+r$, from (\ref{restricted-eqn2}), it follows that $|h_{1}\,\hat{}A| = h_{1}(k-h_{1})+1$. If $h_{1}\geq 2$, then by Theorem \ref{theoremII}, the set $A$ is an arithmetic progression. Therefore, let $h_{1}=1$.
	
	If $h_{2}\geq 3$, then $h_{2}-h_{1}\geq 2$. By (\ref{restricted-eqn2}), we get $|S_{2}|=|(h_{2}-h_{1})\,\hat{}A_{2}|=(h_{2}-h_{1})(k-h_{2})+1$, where $A_{2}=\{a_{1},a_{2},\ldots,a_{k-1}\}$. Therefore, by Theorem \ref{theoremII}, the set $A_{2}$ is an arithmetic progression. To show that $A$ is an arithmetic progression, it is left to show that $a_{k}-a_{k-1}=a_{k-1}-a_{k-2}$. Consider the following integers:
	\begin{multline*}
	a_{k-2} &lt; a_{1}+\cdots+a_{h_{2}-1}+a_{k-2} &lt; a_{1}+\cdots+a_{h_{2}-1}+a_{k-1} &lt; a_{1}+\cdots+a_{h_{2}-1}+a_{k}\\
	= \min(S_{2}).
	\end{multline*}	
	But we already have
	\[a_{k-2}&lt;a_{k-1}&lt;a_{k}&lt;a_{1}+\cdots+a_{h_{2}-1}+a_{k}=\min(S_{2}),\]
	where $\{a_{k-2}, a_{k-1}, a_{k}\} \subseteq S_{1}$.
	Thus
	\[a_{1}+\cdots+a_{h_{2}-1}+a_{k-2}=a_{k-1} \ \text{and} \ a_{1}+\cdots+a_{h_{2}-1}+a_{k-1}=a_{k}.\]
	This implies
	\[a_{k}-a_{k-1}=a_{1}+\cdots+a_{h_{2}-1}=a_{k-1}-a_{k-2},\]
	and we are done.
	
	
	Now let $h_{2}=2$; i.e., $h_{2}-h_{1}=1$. Set $T_{1} := \{a_{1}, a_{2}, a_{1}+a_{2}, a_{1}+a_{3}, \ldots, a_{1}+a_{k-1}\}$.
	Clearly $T_{1} \subseteq h_{1}\,\hat{}A \cup h_{2}\,\hat{}A \subseteq H\,\hat{}A$ and $\max(T_{1})=a_{1}+a_{k-1}&lt;a_{1}+a_{k}=\min(S_{2})$. Therefore $T_{1}=S_{1}$. That is 
	\[\{a_{1}, a_{2}, a_{1}+a_{2}, a_{1}+a_{3}, \ldots, a_{1}+a_{k-1}\}=\{a_{1}, a_{2}, a_{3}, \ldots, a_{k}\}.\]
	Thus $a_{i}=a_{1}+a_{i-1}$ for $i=3,4,\ldots, k$. Equivalently, $a_{i}-a_{i-1}=a_{1}$ for $i=3,4,\ldots, k$. To show that $A$ is an arithmetic progression it is enough to show $a_{2}-a_{1}=a_{k}-a_{k-1}$. Consider the integer $a_{2}+a_{k-1}$. Since $\max(S_{1})=\max(T_{1})=a_{1}+a_{k-1}&lt;a_{2}+a_{k-1}&lt;a_{2}+a_{k}=\min\nolimits^{+} (S_{2})$ and $\max(S_{1})=\max(T_{1})=a_{1}+a_{k-1}&lt;a_{1}+a_{k}=\min(S_{2})&lt;a_{2}+a_{k}=\min\nolimits^{+} (S_{2})$, we must have $a_{2}+a_{k-1}=a_{1}+a_{k}$. This proves $A$ is an arithmetic progression.
	
	
	Next we show that $H$ is an arithmetic progression. For $i=1,2, \ldots,r-1$, consider the following integers:
	\begin{multline*}
	\max\nolimits_{-} (S_{i})
	=a_{k-h_{i}}+a_{k-h_{i}+2}+\cdots+a_{k} &lt; a_{k-h_{i}+1}+\cdots+a_{k}=\max(S_{i}) \\
	&lt; a_{1}+\cdots+a_{h_{i+1}-h_{i}}+a_{k-h_{i}+1}+\cdots+a_{k}=\min(S_{i+1})
	\end{multline*}	
	and
	\begin{multline*}
	\max\nolimits_{-} (S_{i})
	=a_{k-h_{i}}+a_{k-h_{i}+2}+\cdots+a_{k} &lt; a_{1}+\cdots+a_{h_{i+1}-h_{i}}+a_{k-h_{i}}+a_{k-h_{i}+2}+\cdots+a_{k}\\
	&lt; a_{1}+\cdots+a_{h_{i+1}-h_{i}}+a_{k-h_{i}+1}+\cdots+a_{k}=\min(S_{i+1}).
	\end{multline*}
	Therefore \[a_{k-h_{i}+1}+\cdots+a_{k}=a_{1}+\cdots+a_{h_{i+1}-h_{i}}+a_{k-h_{i}}+a_{k-h_{i}+2}+\cdots+a_{k}\]
	for $i=1,2, \ldots, r-1$. This implies
	\begin{equation}\label{equation3.2}
	a_{k-h_{i}+1}-a_{k-h_{i}}=a_{1}+\cdots+a_{h_{i+1}-h_{i}} \ \text{for} \ i=1,2, \ldots,r-1.
	\end{equation}
	Since $A$ is an arithmetic progression, the difference between any two consecutive elements in $A$ is same. Therefore \[a_{2}-a_{1}=a_{k-h_{i}+1}-a_{k-h_{i}}=a_{1}+a_{2}+\cdots+a_{h_{i+1}-h_{i}} \ \text{for} \ i=1,2, \ldots,r-1.\] 
	This holds, only if $h_{i+1}-h_{i}=1$ for $i=1,2, \ldots,r-1$. Hence, $H=h_{1}+[0, r-1]$ and $A=a_{1} \cdot [1, k]$. This completes the proof of the theorem.
	
\end{proof}


\begin{corollary}\label{restricted-sumset-inverse-cor}
	Let $A$ be a set of $k\geq 7$ nonnegative integers with $0 \in A$. Let $H=\{h_{1}, h_{2}, \ldots, h_{r}\}$ be a set of $r\geq 2$ positive integers with $h_{1}&lt;h_{2}&lt;\cdots&lt;h_{r}\leq k-1$. Set $h_{0}=0$. If
	\[|H\,\hat{}A|=\sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-h_{i}-1)+r+1,\]
	then $H=h_{1}+[0, r-1]$ and 
	$A=\min(A\setminus \{0\}) \cdot [0, k-1]$.
\end{corollary}

\begin{proof}
	Let $A^{\prime} := A \setminus \{0\}$. Then the equality $|H\,\hat{}A| = \sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-h_{i}-1)+r+1$ implies $|H\,\hat{}A^{\prime}| = \sum_{i=1}^{r} (h_{i}-h_{i-1}) (k-1-h_{i})+r$. Thus by Theorem \ref{restricted-sumset-inverse-thm}, we get $H=h_{1}+[0, r-1]$ and $A^{\prime} = \min(A^{\prime}) \cdot [1, k-1]$. Hence, $H=h_{1}+[0, r-1]$ and $A=\min(A^{\prime}) \cdot [0, k-1]$. This proves the theorem.
\end{proof}


The following inverse result (which has recently been proved) is a particular case of Theorem \ref{restricted-sumset-inverse-thm} and Corollary \ref{restricted-sumset-inverse-cor}.

\begin{corollary} {\cite[Theorem 2.2, Corollary 2.3]{bhanja}}
	Let $A$ be a set of $k\geq 7$ nonnegative integers and $H=[0, r]$ with $2\leq  r\leq k-1$. If $0 \notin A$ and $|H\,\hat{}A| = rk-\frac{r(r-1)}{2}+1$, then $A = d \cdot [1, k]$ for some positive integer $d$.

If $0 \in A$ and $|H\,\hat{}A| = rk-\frac{r(r+1)}{2}+1$, then $A = d \cdot [0, k-1]$ for some positive integer $d$.
\end{corollary}


\section{Acknowledgement}
We would like to thank Professor R. Thangadurai for carefully going
through this manuscript. We would also like to thank the editor-in-chief
and the anonymous referee for the helpful comments and suggestions which
helped improve the article. We also acknowledge the support of the PDF
scheme of Harish-Chandra Research Institute, Prayagraj.


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\bibitem{bajnok} B. Bajnok, {\it Additive Combinatorics: A Menu of Research Problems}, Discrete Mathematics and its Applications, CRC Press, 2018.

\bibitem{balandraud} \'{E}. Balandraud, Addition theorems in $\mathbb{F}_{p}$ via the polynomial method, preprint, 2017. Available at \url{https://arxiv.org/abs/1702.06419}.

\bibitem{bhanja} J. Bhanja and R. K. Pandey, Inverse problems for certain subsequence sums in integers, {\it Discrete Math.} {\bf 343} (2020), 112148.

\bibitem{nathu} M. B. Nathanson, Inverse theorems for subset sums, {\it Trans. Amer. Math. Soc.} {\bf 347} (1995), 1409--1418.

\bibitem{nathanson} M. B. Nathanson, {\it Additive Number Theory: Inverse Problems and the Geometry of Sumsets}, Springer, 1996.

\end{thebibliography}

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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11P70; Secondary 11B75, 11B13.

\noindent \emph{Keywords:} sumset, restricted sumset.

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\vspace*{+.1in}
\noindent
Received August 22 2020;
revised versions received  August 23 2020; February 1 2021; March 27 2021;
March 28 2021.
Published in {\it Journal of Integer Sequences}, March 28 2021.

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